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The following notion is introduced by Assaf Rinot:

Definition. A singular cardinal $\kappa$ is a prevalent singular cardinal iff there exists a family $\mathbb{A}\subset P(\kappa)$ with $|\mathbb{A}| = \kappa$ and $sup\{|A| : A\in \mathbb{A}\} < \kappa$ such that any $B\subset \kappa$ with $|B| < cf(\kappa)$ is contained in some $A\in \mathbb{A}$.

Prevalent Singular Cardinals Hypothesis (PSCH) states that any singular cardinal is a prevalent singular cardinal.

Clearly any singular cardinal of countable cofinality is prevalent.

Question. What is known about the consistency of the failure of PSCH?

Remark 1. PSCH follows from $GCH, SCH, PFA$ and many other combinatorial principles.

Remark 2. As stated in Assaf Rinot's paper "On topological spaces of singular density and minimal weight, Topology and its Applications 155 (2007) 135–140", $PSCH$ is a very weak assertion and all currently known methods for violating statements of similar flavor, will fail to violate the $PSCH$.

Question 2. Is there any relation between PSCH and some combinatorial principles introduced by Shelah in PCF theory like $Cov$ or ...?

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  • $\begingroup$ Just to see that I got it, if $2^\omega=\aleph_{\omega_1}$ then it is not prevalent. Right? $\endgroup$ – Asaf Karagila Nov 26 '13 at 10:38
  • $\begingroup$ I don't understand what you mean.but let me mention the following: Start with $GCH$ and add $\aleph_{\omega_1}-$many Cohen reals. In the extension $\aleph_{\omega_1}$ is prevalent: just consider $([\aleph_{\omega_1}]^{<\omega_1})^V$ as your family and note that any countable set of ordinals in the extension is covered by a countable ground model set. $\endgroup$ – Mohammad Golshani Nov 26 '13 at 10:46
  • $\begingroup$ Ah, I see. I somehow read that the covering family should have size $<\kappa$ and not $\kappa$. But if we add $\aleph_{\omega_1+1}$ Cohen reals, then it is no longer prevalent. $\endgroup$ – Asaf Karagila Nov 26 '13 at 11:45
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    $\begingroup$ No, still the above argument works. In fact as I stated in the post, $SCH$ implies $PSCH$ and so you can not force the failure of $PSCH$ by these simple arguments. $\endgroup$ – Mohammad Golshani Nov 26 '13 at 11:56
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    $\begingroup$ As stated in some papers by Assaf, $PSCH$ holds in all known models. $\endgroup$ – Mohammad Golshani Nov 26 '13 at 11:57
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A singular cardinal $\lambda$ is prevalent iff there exists some cardinal $\mu<\lambda$ such that $Cov(\lambda,\mu,cf(\lambda),2)=\lambda$. In his solution of the pcf conjecture, Gitik has constructed a model where there exists a singular cardinal $\lambda$ of cofinality $\aleph_1$ such that $pp(\mu)>\lambda$ for cofinally many $\mu<\lambda$ of countable cofinality. Given $\mu<\lambda$, pass to a bigger $\mu'<\lambda$ of countable cofinality with $pp(\mu')>\lambda$. Then, we have $pp(\mu')\le Cov(\mu',\mu',cf(\mu')^+,2)\le Cov(\lambda,\mu',cf(\lambda),2)\le Cov(\lambda,\mu,cf(\lambda),2)$. So $Cov(\lambda,\mu,cf(\lambda),2)>\lambda$ for every $\mu<\lambda$, meaning that $\lambda$ is not prevalent.

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  • $\begingroup$ Interesting, thanks so much for your answer. $\endgroup$ – Mohammad Golshani Nov 27 '13 at 12:04
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A singular cardinal $\kappa$ is prevalent iff $\operatorname{cov} (\kappa , \mu , \operatorname{cf} (\kappa)) = \kappa$ for some cardinal $\mu$ with $\operatorname{cf} (\kappa) \leq \mu < \kappa$.

Proof: First, it is easy to show that $\operatorname{cov} (\kappa , \mu , \theta) \geq \kappa$ when $\theta \leq \mu < \kappa$.
Now, if $\kappa$ is prevalent, then $\operatorname{cov} (\kappa , \mu , \operatorname{cf} (\kappa)) \leq \kappa$, with $\eta = \operatorname{sup} \{ |A| : A \in \mathbb{A} \}$ and $$ \mu = \left\lbrace \begin{array}{ll} \eta , & \textrm{if } |A| < \eta \textrm{ for all } A \in \mathbb{A} ; \\ \eta^{+} , & \textrm{otherwise.} \end{array} \right. $$ The other direction is immediate. $\square$

SSH (Shelah's Strong Hypothesis) implies SCH, hence implies PSCH. By Corollary 3.6 in

Pierre Matet. Large cardinals and covering numbers, Fundamenta Mathematicae 205 (2009), 45-75. doi:10.4064/fm205-1-3

SSH is equivalent to the following: given cardinals $\mu$ and $\lambda$, with $\mu \geq \lambda = \operatorname{cf} (\lambda) \geq \aleph_{1}$, $$ \operatorname{cov} \left( \mu , \lambda , \lambda \right) = \left\lbrace \begin{array}{ll} \mu , & \textrm{if } \operatorname{cf} (\mu) \geq \lambda ; \\ \mu^{+} , & \textrm{otherwise.} \end{array} \right. $$

Under SSH, when $\kappa$ is a singular cardinal with $\operatorname{cf} (\kappa) > \aleph_0$, we have, for any $\mu$ with $\operatorname{cf} (\kappa) \leq \mu < \kappa$: $$ \operatorname{cov} (\kappa , \mu , \operatorname{cf} (\kappa)) \leq \operatorname{cov} (\kappa , \operatorname{cf} (\kappa) , \operatorname{cf} (\kappa)) = \kappa . $$

Note that the condition (weaker than prevalent) $\operatorname{cov} (\kappa , \kappa , \operatorname{cf} (\kappa)) \leq \kappa$ (for a singular cardinal $\kappa$) is a theorem in ZFC. In fact, $\operatorname{cov} (\kappa , \kappa , \operatorname{cf} (\kappa)) = \operatorname{cf} (\kappa)$, by Observation 5.2(3) in Chapter II of

S. Shelah. Cardinal Arithmetic, volume 29 of Oxford Logic Guides. Oxford University Press, New York, 1994.

See Definition 5.1 in Chapter II for $\operatorname{cov} (\lambda , \kappa , \theta , \sigma)$. Then, $\operatorname{cov} (\lambda , \kappa , \theta) := \operatorname{cov} (\lambda , \kappa , \theta , 2)$.

Finally, the condition "$|B| < \operatorname{cf} (\kappa)$" in the definition of "prevalent singular cardinal" cannot be replaced by "$|B| \leq \operatorname{cf} (\kappa)$", because, for any singular cardinal $\kappa$ and any $\mu$ with $\operatorname{cf} (\kappa) < \mu < \kappa$ we have $$ \operatorname{cov} (\kappa , \mu , {(\operatorname{cf} (\kappa))}^+) \geq \operatorname{cov} (\kappa , \kappa , {(\operatorname{cf} (\kappa))}^+) \geq \operatorname{cov} (\kappa , \kappa , {(\operatorname{cf} (\kappa))}^+ , \operatorname{cf} (\kappa)) \geq {\kappa}^+ , $$ by Fact 1 in

Andreas Liu, Bounds for covering numbers, The Journal of Symbolic Logic 71 (2006), 1303-1310. doi:10.2178/jsl/1164060456

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